Question

calculate how much a 5.6 m slab of glass will expand if heated from -79°c to 68°c. the coefficient of thermal expansion for glass is 9×10^-6 1/°c. express your answer in m, cm, & in. 1 in = 2.54 cm.

1. -/1 points

calculate how much heat is required to bring 355 g of water from a temperature of 72.7°c to a temperature of 100°c. the specific heat of water is 4186 j/kg·°c

Explanation:

First Problem (Thermal Expansion of Glass)

Step1: Recall the thermal expansion formula

The formula for linear thermal expansion is $\Delta L = \alpha L_0 \Delta T$, where $\Delta L$ is the change in length, $\alpha$ is the coefficient of thermal expansion, $L_0$ is the original length, and $\Delta T$ is the change in temperature.

Step2: Calculate the change in temperature

$\Delta T = T_2 - T_1 = 68^\circ\text{C} - (-79^\circ\text{C}) = 68 + 79 = 147^\circ\text{C}$

Step3: Substitute values into the expansion formula (in meters)

Given $\alpha = 9\times 10^{-6} \frac{1}{^\circ\text{C}}$, $L_0 = 5.6\ \text{m}$, $\Delta T = 147^\circ\text{C}$
$\Delta L = (9\times 10^{-6}) \times 5.6 \times 147$
First, calculate $5.6 \times 147 = 823.2$
Then, $\Delta L = 9\times 10^{-6} \times 823.2 = 7.4088\times 10^{-3}\ \text{m}$

Step4: Convert meters to centimeters

Since $1\ \text{m} = 100\ \text{cm}$, $\Delta L_{\text{cm}} = 7.4088\times 10^{-3}\ \text{m} \times 100 = 0.74088\ \text{cm}$

Step5: Convert centimeters to inches

Given $1\ \text{in} = 2.54\ \text{cm}$, $\Delta L_{\text{in}} = \frac{0.74088\ \text{cm}}{2.54\ \text{cm/in}} \approx 0.2917\ \text{in}$

Second Problem (Heat Required for Water)

Step1: Recall the heat formula

The formula for heat transfer is $Q = mc\Delta T$, where $Q$ is the heat, $m$ is the mass, $c$ is the specific heat, and $\Delta T$ is the change in temperature.

Step2: Convert mass to kilograms

Given $m = 355\ \text{g} = 0.355\ \text{kg}$

Step3: Calculate the change in temperature

$\Delta T = 100^\circ\text{C} - 72.7^\circ\text{C} = 27.3^\circ\text{C}$

Step4: Substitute values into the heat formula

$Q = 0.355\ \text{kg} \times 4186\ \text{J/kg}^\circ\text{C} \times 27.3^\circ\text{C}$
First, calculate $0.355 \times 4186 = 1486.03$
Then, $Q = 1486.03 \times 27.3 \approx 40568.62\ \text{J}$ (or more precisely, $0.355\times4186 = 1486.03$; $1486.03\times27.3 = 1486.03\times27 + 1486.03\times0.3 = 40122.81 + 445.809 = 40568.619\ \text{J}$)

Answer:

s:

First Problem:

• m: $\boldsymbol{7.41\times 10^{-3}\ \text{m}}$ (or $0.00741\ \text{m}$)
• cm: $\boldsymbol{0.741\ \text{cm}}$ (rounded)
• in: $\boldsymbol{0.292\ \text{in}}$ (rounded)

Second Problem:

• J: $\boldsymbol{\approx 4.06\times 10^4\ \text{J}}$ (or $40569\ \text{J}$)